Is the derivative the natural logarithm of the left-shift?
Well, it seems that you have just discovered a beautiful theory of (semi)group generators by yourself. To give some basics of it, let us consider a collection of "nice" functions on real values - e.g. bounded and having continuous derivatives. The action of operators $L^h$ on this space has a semigroup structure: $$ L^s(L^tf(x)) = L^sf(x+t) = f(x+s+t) = L^{s+t}f(x). \tag{1} $$ Also, you have that $L^0f(x) = f(x)$, so $L^0$ is the identity operator -it does not change its argument. You can see, that although there are a lot of operators in the collection $(L^h)_{h}$, they have to satisfy the semigroup property $L^{s+t} = L^sL^t$, and so there is no much freedom in choosing them. Even more, one can define the generator of the semigroup (also sometimes called the derivative of it) by $$ \mathscr Af(x):=\lim_{h\to 0}\frac{1}{h}(L^hf - f) $$ which in your case exactly coincides with the derivative of the function. However, if you would consider the semigroup $K^hf(x) = f(x + v\cdot h)$ for some constant $v$, you'll see that the generator will be a bit different. Anyways, under certain condition - if you don't know the semigroup $L$, but you're just given the generator $\mathscr A$, it is possible to reconstruct $L$ from $\mathscr A$ by the so-called exponential map, that is $$ L^h:=\mathrm e^{h\mathscr A} $$ where the definition of the exponent of the operator indeed is given by the Taylor series where e.g. $\mathscr A^2 f(x) = \mathscr A(\mathscr A f(x))$. As a result, you can indeed consider $\mathscr A$ to be a certain logarithm of $L^h$ and it comes as no surprise that it has a similar Taylor expansion. However, although the name "exponential map" from $\mathscr A$ to $L^h$ is commonly used, I haven't heard of the inverse being called the "logarithm". One rather uses the "generator" or "derivative." If you are further interested, I would suggest you reading the linked wikipedia article.
$$ \begin{align} e^{d/dx} x^n = {} & \left(1+\frac{d}{dx} + \frac{(d/dx)^2}{2}+ \frac{(d/dx)^3}{6}+\frac{(d/dx)^4}{24}+\cdots+\frac{(d/dx)^k}{k!}+\cdots\right) x^n \\[10pt] = {} & x^n + nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3}+\cdots \\[10pt] & \cdots+\frac{n(n-1)\cdots(n-k+1)}{k!}x^{n-k}+\cdots+0+0+0+\cdots \\[10pt] = {} & \binom n0 x^n 1^0 + \binom n1 x^{n-1} 1^1 + \binom n2 x^{n-2} 1^2 + \cdots + \binom n n x^0 1^n \\[10pt] & \cdots+0+0+0+\cdots \\[10pt] = {} & (x+1)^n. \end{align} $$ So the binomial theorem entails that $e^{d/dx} = L$ at least as applied to polynomials.
If applied to functions other than polynomials, there are convergence issues to examine.
how does one make sense of exponentiating or taking the logarithm of an operator?
The operator is linear, and therefore so are its positive integer powers, hence any power series in that operator has a chance of making sense. At least the series is a limit of linear operators, and the series makes perfect sense without any limiting process when applied to polynomials, because almost all terms are zero.
In the 17-18th century, functions were believed to have power series representations, so that formal manipulations that were provably correct for polynomials ought to somehow justify the same calculation in general. This is not quite right but it works spectacularly as a heuristic.
What, if any, significance does a statement like d/dx=ln L have?
It adds precision to the analogy between calculus and the "calculus of finite differences". As you might guess, what is true for differentiation extends to integration. The corresponding relationship between finite sums and integrals is the Euler-Maclaurin summation formula and it is derived by taking the reciprocal of a power series equivalent to the one you wrote down.